How to make a bounded set A in R2 with Hausdorff dimension of 1 but $H_{1}(A)$ = $\infty$?

Question

I have been tasked with creating a set in $A \subset \mathbb{R}^2$ such that $A$ is bounded and has Hausdorff dimension of 1 but $H_{1}(A) = \infty$.

I started by creating a bounded space, working in the unit square centered at the origin. I then created a circle centered at the origin with radius 1 so that it touches the center of each side of the square but leaves room. I then continue by drawing circles in the space that remains. I would then continue this procedure, eventually getting an infinite amount of circles with decreasing radius all within the unit square. The motivation for this was to create an infinite sequence of circles that have radii under some arbitrary delta, but which ultimately sum up to infinity. However, I think that this procedure will be unsuccessful, since such fractal like images tend to have Hausdorff dimension greater than 1. I know that the Hausdorff dimension is the limit as delta goes to $0$ of the infimum of the sum of the diameter of the sets of covers of the set $A$ where each set in the cover has diameter less than delta.

How should one approach the problem of constructing this set $A$? I assume fractal patterns will not be the way to go. The requirement of making $A$ bounded makes the problem much more difficult.

Answer 1

The basic result is:

A countable union of $d$-dimensional sets is still $d$-dimensional.

Indeed, the Hausdorff dimension of a set $E$ is defined as $$\text{dim}(E)=\inf\,\left\{s>0:\mathcal{H}^s(E)=0\right\}.$$ If you take a countable family $\{E_i\}$ of $d$-dimensional sets, by definition for every $t>s$ we have $\mathcal{H}^t(E_i)=0$. Since $\mathcal H^t$ is an outer measure, by $\sigma$-subadditivity, setting $E=\bigcup_i E_i$, we have $\mathcal{H}^t(E)=0$, from which $\text{dim}(E)\leq s$ (if your restrict to Borel sets you can just use $\sigma$-additivity, because $\mathcal{H}^t$ is a measure on the Borel sets). The other inequality is trivial.

With the same proof you obtain

Given a countable family $\{E_i\}$ and setting $E=\bigcup_i E_i$, then $$\text{dim}(E)=\sup_i\, \text{dim}(E_i).$$

Therefore your example is of Hausdorff dimension $1$, but the delicate part is to show that the total length diverges! For that you can have a look at this question (if I understood correctly your proposed construction). Anyway you can build a simpler example by taking concentric circles whose radii add up to $\infty$, or considering the example proposed by Silvia Ghinassi in the comments.

Regarding fractals, it is true that taking a limiting procedure the dimension of the limit object could be greater than the dimension of the approximating sets: for example the Koch snowflake is a limit of $1$-dimensional curves, but it is of dimension $\log 4/\log 3$. However, this can not happen just by taking a countable union; a limiting process is really required!